Exponential Function as an Infinite Product

Is there any representation of the exponential function as an infinite product (where there is no maximal factor in the series of terms which essentially contributes)? I.e.

$$\mathrm e^x=\prod_{n=0}^\infty a_n,$$

and by the sentence in brackets I mean that the $a_n$'s are not just mostly equal to $1$ or pairwise canceling away. The product is infinite but its factors don't contain a subseqeunce of $1$, if that makes sense.

There is of course the limit definition as powers of $(1+x/n)$., but these are no definite $a_n$'s, which one could e.g. divide out.

Not sure if this satisfies your assumptions, but this is an interesting infinite product for $|z|<1$, $$e^z=\prod_{k=1}^\infty (1-z^k)^{-\frac{\mu(k)}{k}},$$ where $\mu(k)$ is the Möbius function. See here.

Proof

Let's start with the logarithm of the product, $$-\sum_{k=1}^\infty\frac{\mu(k)}{k}\log\left(1-z^k\right)=\sum_{k=1}^\infty\frac{\mu(k)}{k}\sum_{\ell=1}^\infty\frac{z^{k\ell}}{\ell}.$$ We can rewrite this $$\sum_{n=1}^\infty\frac{z^n}{n}\sum_{d\mid n}\mu(d)=z+\sum_{n=2}^\infty\frac{z^n}{n}\sum_{d\mid n}\mu(d)=z$$ since $\sum_{d\mid n}\mu(d)=0$ for $n\geq 2$.

There exists an infinite product for $e$ as follows:

If we define a sequence $\lbrace e_n\rbrace$ by $e_1=1$ and $e_{n+1}=(n+1)(e_n+1)$ for $n=1,2,3,...;$ e.g. $$e_1=1,e_2=4,e_3=15,e_4=64,e_5=325,e_6=1956,...$$ then $$e=\prod_{n=1}^\infty\frac{e_n+1}{e_n}=\frac{2}{1}.\frac{5}{4}.\frac{16}{15}.\frac{65}{64}.\frac{326}{325}.\frac{1957}{1956}. ...$$ For proof, first by induction we can show that if $s_n=\sum_{k=0}^n\frac1 {k!}$, then $e_n=n!s_{n-1}$,for $n\in\mathbb N$. And this immediately follows that $s_n/s_{n-1}=(e_n+1)/e_n$ and $s_n=\prod_{k=1}^n\frac{e_k+1}{e_k}$. Hence, $$e^x=\prod_{n=1}^\infty\left(\frac{e_n+1}{e_n}\right)^x.$$

If $x\geqslant0$ (or $x\ne-2^n$ for every $n\geqslant0$), one can use $$a_0=1+x,\qquad a_{n+1}=\left(1+\frac{x^2}{2^{n+2}(x+2^n)}\right)^{2^n} $$ If $x\leqslant0$ (or $x\ne2^n$ for every $n\geqslant0$), one can use $$a_0=\frac1{1-x},\qquad a_{n+1}=\left(1-\frac{x^2}{(2^{n+1}-x)^2}\right)^{2^n} $$ Where does this come from? From the identity, valid for every $n\geqslant0$, $$ \prod_{k=0}^na_k=\left(1\pm\frac{x}{2^n}\right)^{\pm2^n}. $$ The first identity (when $\pm=+$) yields a nondecreasing sequence of partial products. The second identity (when $\pm=-$) yields a nonincreasing sequence of partial products.

Amazingly, the exponential function can be represented as an infinite product of a product! That result was shown in the 2006 paper "Double Integrals and Infinite Products For Some Classical Constants Via Analytic Continuations of Lerch's Transendent" by Jesus Guillera and Jonathan Sondow.

It is proven in Theorem 5.3 that $$e^x=\prod_{n=1}^\infty \left(\prod_{k=1}^n (kx+1)^{(-1)^{k+1} {{n}\choose{k}}}\right) ^{1/n}$$

I dunno, this was too cool not to show you.